Find them, Catch them
 
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 32225   Accepted: 9947

Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:

1. D [a] [b] 
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

2. A [a] [b] 
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang. 

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

Sample Input

1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4

Sample Output

Not sure yet.
In different gangs.
In the same gang. 题目大意:N 表示共有 N 个帮派;下面有 M 条个询问
有两个帮派,现在告诉你一些互相敌对的点对,然后让你判断某两个点之间的关系,并首先判两个点关系是否确定;
用并查集的思想只要两个点在一个集合里就可以了,直接用并查集然后时候在同一个集合;
AC代码:
 #include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
// 0 代表木有关系,1表示不同类,2代表同类
const int N = 1e5+;
int m;
int f[N],c[N];
int init()
{
for(int i=; i<=m; i++)
{
f[i] = i;
c[i] = ;
}
}
int xfind(int x)
{
if(f[x] == x)
return x;
int t = f[x];
f[x] =xfind(f[x]);
c[x] = (c[x]+c[t])%;
return f[x];
}
int solve(char ch,int x,int y )
{
int fx = xfind(x);
int fy = xfind(y);
if(fx != fy)
{
if(ch == 'A') return ; //返回值0,表示不确定
f[fx] =fy;
if((c[x]+c[y])% ==)
c[fx] = ;
}
else
{
if(ch == 'A')
{
if(c[x] != c[y]) return ; //返回值1,表示在不同的帮派
if(c[x] == c[y]) return ; //返回值2,表示在相同的帮派
}
}
if(ch == 'D') return ;//这种情况不同在考虑的
}
int main( )
{
int T,n,x,y;
char s;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&m,&n);
init(); //初始化并查集
for(int i=; i<n; i++)
{
scanf(" %c %d %d",&s,&x,&y);
int flag = solve(s,x,y);
if(flag == ) printf("Not sure yet.\n");
else if(flag == ) printf("In different gangs.\n");
else if(flag == ) printf("In the same gang.\n");
}
}
return ;
}

poj1703 Find them, Catch them(并查集的应用)的更多相关文章

  1. poj1703 Find them, Catch them 并查集

    poj(1703) Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26992   ...

  2. POJ-1703 Find them, Catch them(并查集&数组记录状态)

    题目: The police office in Tadu City decides to say ends to the chaos, as launch actions to root up th ...

  3. POJ 1703 Find them, catch them (并查集)

    题目:Find them,Catch them 刚开始以为是最基本的并查集,无限超时. 这个特殊之处,就是可能有多个集合. 比如输入D 1 2  D 3 4 D 5 6...这就至少有3个集合了.并且 ...

  4. poj1703--Find them, Catch them(并查集应用)

    Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 32073   Accepted: ...

  5. POJ1703-Find them, Catch them 并查集构造

                                             Find them, Catch them 好久没有做并查集的题,竟然快把并查集忘完了. 题意:大致是有两个监狱,n个 ...

  6. POJ 2236 Wireless Network ||POJ 1703 Find them, Catch them 并查集

    POJ 2236 Wireless Network http://poj.org/problem?id=2236 题目大意: 给你N台损坏的电脑坐标,这些电脑只能与不超过距离d的电脑通信,但如果x和y ...

  7. POJ 1703 Find them, Catch them 并查集的应用

    题意:城市中有两个帮派,输入中有情报和询问.情报会告知哪两个人是对立帮派中的人.询问会问具体某两个人的关系. 思路:并查集的应用.首先,将每一个情报中的两人加入并查集,在询问时先判断一下两人是否在一个 ...

  8. poj1703_Find them, Catch them_并查集

    Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42451   Accepted: ...

  9. poj.1703.Find them, Catch them(并查集)

    Find them, Catch them Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I6 ...

  10. POJ 1703 Find them, Catch them(并查集高级应用)

    手动博客搬家:本文发表于20170805 21:25:49, 原地址https://blog.csdn.net/suncongbo/article/details/76735893 URL: http ...

随机推荐

  1. Openshift 用户,角色和RBAC

    OCP中的权限管理沿用的Kubernetes RBAC机制,授权模式主要取决于下面几个因数 Rules 针对主要对象的操作权限,比如建立Pod Sets of permitted verbs on a ...

  2. phpMyAdmin配置

    一: (1):下载phpmyadmin,在官方最好               (2):个人建议最好将其安装在apache的htdocs文件中(apache的默认虚拟目录,或web访问目录)      ...

  3. Tomcat集群环境下session共享方案 通过memcached 方法实现

    对于web应用集群的技术实现而言,最大的难点就是:如何能在集群中的多个节点之间保持数据的一致性,会话(Session)信息是这些数据中最重要的一块.要实现这一点, 大体上有两种方式:一种是把所有Ses ...

  4. 转:关于Android机型适配这件小事儿

    http://www.umindex.com/devices/android_resolutions 大家都知道Android机型分裂严重,在开发Android App的时候永远都面临适配N多机型的问 ...

  5. Office WPS如何让页与页之间不相互影响

    在一个页面结束的位置点击插入-分页符,完了之后测试按回车下一页的内容有没有跟着往下跑,如果还是跟着往下跑的,再插入一次分页符,一般插入多次之后就不会跟着跑了,但是插的太多会有空白页面   测试按回车, ...

  6. IM开发基础知识补课(四):正确理解HTTP短连接中的Cookie、Session和Token

    本文引用了简书作者“骑小猪看流星”技术文章“Cookie.Session.Token那点事儿”的部分内容,感谢原作者. 1.前言 众所周之,IM是个典型的快速数据流交换系统,当今主流IM系统(尤其移动 ...

  7. python fabric使用 http://fabric-chs.readthedocs.io/zh_CN/chs/tutorial.html

    fab -u username -p password  -H hostname -P -- cmd   或root@'hostname'  -H多个主机是引号用逗号隔开 -P异步

  8. 无埋点数据收集和adb monkey测试屏蔽通知栏

    简单记录百度移动统计android无埋点sdk使用和monkey测试屏蔽通知栏的问题 1.无埋点sdk使用 很简单,下载完sdk后导入到项目中 , 参考sdk文档进行就可以了,个人觉得比友盟还简单,几 ...

  9. 我的Android进阶之旅------&gt;Android字符串资源中的单引號问题error: Apostrophe not preceded by 的解决的方法

    刚刚在string字符串资源文件里,写了一个单引號.报错了,错误代码例如以下 error: Apostrophe not preceded by \ (in OuyangPeng's blog ) 资 ...

  10. 《暗黑世界V1.4》API说明文档

    <暗黑世界V1.4>API说明文档 阵法位置示意图 上方:                        下方:      账号注册   100 请求信息 { username   str ...